GC Oxidizing agent

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UW09

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Which substance behaves as the oxidizing agent in
Pb + PbO2+ 2H2SO4 --> 2PbSO4 + 2H20


I am guessing I am doing the oxidation numbers wrong?

(0) + (+4)(-2) + (+1)(+6)(-2) --> (+10)(-2)(-2) + (0)

Answer is PbO2...would appreciate it if someone could explain this.

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Which substance behaves as the oxidizing agent in
Pb + PbO2+ 2H2SO4 --> 2PbSO4 + 2H20


I am guessing I am doing the oxidation numbers wrong?

(0) + (+4)(-2) + (+1)(+6)(-2) --> (+10)(-2)(-2) + (0)

Answer is PbO2...would appreciate it if someone could explain this.
What's the oxidation # for Pb in PbO2?
Pb + 2(-2) = 0-->Pb = +4

What's the oxidation # for Pb in PbSO4?
Pb = +2

The oxidation# has change from +4 to +2, and that means gain of 2e-. Thus, Pb in PbO2 is reduced (gain of electron). Therefore, PbO2 is the oxidizing agent!
 
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What's the oxidation # for Pb in PbO2?
Pb + 2(-2) = 0-->Pb = +4

What's the oxidation # for Pb in PbSO4?
Pb = 2+ (Charge of Pb)

The oxidation# has change from +4 to +2, and that means gain of 2e-. Thus, Pb in PbO2 is reduced (gain of electron). Therefore, PbO2 is the oxidizing agent!

Could you tell me how you went about getting the oxidation number for Pb in PbSO4? I went by the rule of the most electronegative atom getting their typical oxidation state (so O was assigned -2 (total -8)...then S was assigned -2...and finally Pb was given +10 to balance the molecule). Thanks!
 
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Could you tell me how you went about getting the oxidation number for Pb in PbSO4? I went by the rule of the most electronegative atom getting their typical oxidation state (so O was assigned -2 (total -8)...then S was assigned -2...and finally Pb was given +10 to balance the molecule). Thanks!
OK...
I got confused for a second and thought I did this wrong. Sorry for the multiple edits.
Here's how you can do it:

-Figure out oxidation# of S in sulfate SO4^2-

S + 4(-2) = -2 -->S - 8 = -2 --> S = +6

-Figure out the oxidation # of Pb:
PbSO4: Pb + 6 + 4(-2) = 0 --> Pb + 6 - 8 = 0 --> Pb = +2

The easier way is just to assume PbSO4 ionizes:

PbSO4-->Pb2+ + SO4^2-

Then oxidation # of Pb is equal to it's charge = +2

Hope this helps!

 
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Could you tell me how you went about getting the oxidation number for Pb in PbSO4? I went by the rule of the most electronegative atom getting their typical oxidation state (so O was assigned -2 (total -8)...then S was assigned -2...and finally Pb was given +10 to balance the molecule). Thanks!

You need to understand that the compound is a salt...

Pb++ and SO4--

By knowing that sulfate has a -2 charge you can automatically deduce the oxidation state of lead.

Also, the oxidation state of sulfur in sulfate is +6.

The oxidation state of sulfur and oxygen must equal the net charge of the sulfate compound.
 
you need to understand that the compound is a salt...

Pb++ and so4--

of course, this is not completely true! According to solubility rules, PbSO4 is not soluble, so it doesn't ionize in water. But you can assume ionization to solve for the oxidation #.

by knowing that sulfate has a -2 charge you can automatically deduce the oxidation state of lead.

Also, the oxidation state of sulfur in sulfate is +6.

The oxidation state of sulfur and oxygen must equal the net charge of the sulfate compound.
;)
 

Ah ha! I didn't say it was a soluble salt! I just see compounds as salts....I mean, for example...potassium ferrocyanide is insoluble; I automatically see potassium and ferrocyanide (complex ion)...thus, I automatically see that iron in ferrocyanide has an oxidation state of +2

Actually, it's an important point that you have brought up regarding the solubility of lead(II)sulfate. Sulfate in the form of sulfuric acid is present because it drives the redox reaction to further completion by removing Pb++ as it forms to generate the insoluble salt.

The equilibrium constant for electrochemical reaction is multiplied by the inverse of the Ksp of Lead(II)sulfate to generate a greater overall equilibrium constant (going further to completion).

Just a little tid-bit to throw out there.
 
Ah ha! I didn't say it was a soluble salt!

Actually, it's an important point that you have brought up regarding the solubility of lead(II)sulfate. Sulfate in the form of sulfuric acid is present because it drives the redox reaction to further completion by removing Pb++ as it forms to generate the insoluble salt.

The equilibrium constant for electrochemical reaction is multiplied by the inverse of the Ksp of Lead(II)sulfate to generate a greater overall equilibrium constant (going further to completion).

Just a little tid-bit to throw out there.
I just wanted to mess with you a little:laugh:
 
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